Question: When a customer places an order at Derin's bakery, there is an $85\%$ probability that the order will include some kind of dessert. One day, $7$ customers place orders at Derin's bakery. Assuming that each order is equally likely to include dessert, what is the probability that at least one order will not include dessert? Round your answer to the nearest hundredth. $P(\text{at least one without dessert})=$
Strategy In this situation it is much easier to calculate the probability of the event we are looking for (at least one order without any dessert) by calculating the probability of its complement (every order includes some dessert), and subtracting from $1$. In other words, we can use this strategy: $P(\text{at least one without dessert})=1-P(\text{all }7\text{ with dessert})$ Calculations $\begin{aligned} &\phantom{=}P(\text{at least one without dessert}) \\\\ &=1-P(\text{all }7\text{ with dessert}) \\ \\ &=1-(0.85)^{7} \\ \\ &\approx 1-0.3206 \\ \\ &\approx 0.6794\end{aligned}$ Answer $P(\text{at least one without dessert}) \approx 0.68$